Total change in entropy, what am I missing?

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The discussion focuses on calculating the total change in entropy of the universe when a 20.0-kg box slides to rest on a rough table, initially moving at 1.0 m/s. The kinetic energy of the box is converted into heat, which is absorbed by the table and the surrounding room at a constant temperature of 293 K. The entropy change for the box is zero, as it does not lose heat, while the table experiences an increase in entropy due to the absorbed heat. The total change in entropy can be calculated using the formula ΔS = Q/T, where Q is the heat transferred and T is the temperature in Kelvin.

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A 20.0 -kg box having an initial speed of 1.0 m/s slides along a rough table and comes to rest.
Estimate the total change in entropy of the universe. Assume all objects are at room temperature (293 K).

I don't understand how to answer this problem. It feels like I wasn't given enough information to solve for Delta S.
 
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asdf188 said:
A 20.0 -kg box having an initial speed of 1.0 m/s slides along a rough table and comes to rest.
Estimate the total change in entropy of the universe. Assume all objects are at room temperature (293 K).

I don't understand how to answer this problem. It feels like I wasn't given enough information to solve for Delta S.
Assume that the kinetic energy of the box is dissipated as heat and that this heat is absorbed by the table and room at a constant temperature (ie the temperature of the table and room does not change).

Hint: Since the box does not lose any heat, there is 0 change in the entropy of the box. However, the table gains heat.

AM
 

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